Dot Product
Academic Resource Center
In This Presentation…
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We will give a definition
Look at properties
See the relationship in projections
Look at vectors in different coordinate systems
Do example problems
Dot Product
Definition:
If a = <a1, a2> and b = <b1, b2>, then the dot product of a and b
is number a · b given by
a · b = a1b1 + a2b2
Likewise with 3 dimensions,
Given a = <a1, a2, a3> and b = <b1, b2 , b3 >
a · b = a1b1 + a2b2 + a3b3
Dot Product
The result of a dot product is not a vector, it is a real number
and is sometimes called the scalar product or the inner
product.
Orthogonal Vectors
Two vectors a and b are orthogonal (perpendicular) if and only
if a · b = 0
Example: The vectors i, j, and k that correspond to the x, y,
and z components are all orthogonal to each other
Examples
Find a · b:
1. Given a = <-4, -5> and b = <2, -9>
a · b = a1b1 + a2b2
a · b = (-4)(2) + (-5)(-9) = -8 + 45
a · b = 37
2.
Given a = 2j + 7k and b = -7i + 4j - k
a · b = (0)(-7) + (2)(4) + (7)(-1) = 0 + 8 + (-7)
a·b=1
Dot Product
Properties of the dot product
1. a · a = |a|2
2. a · b = b · a
3. a · (b + c) = a · b + a · c
4. (ca) · b = c(a · b) = a · (cb)
5. 0 · a = 0
(Note that 0 (bolded) is the zero vector)
Dot Product
If the angle between the two vectors
a and b is θ, then
a · b = |a||b|cos θ
or
θ=
Examples
Find a · b:
1. Given |a| = 8, |b| = 4 and θ = 60°
a · b = |a||b|cos θ = (8)(4)cos(60°) = 12(1/2)
a·b=6
2.
Given |a| = 3, |b| = 2 and θ = π/4
a · b = (3)(2)cos(π/4) = 6(√(2)/2)
a · b = 3√(2)
Example
Find the angle between the two vectors:
Given a = <-1, -3> and b = <3, -3>
a · b = (-1)(3) + (-3)(-3) = -3 + 9 = 6
|a| = √(1+9) = √(10)
|b| = √(9+9) = √(18) = 3√(2)
θ = 6/(√(10)*3√(2)) = 6/(6√(5)) = 1/√(5)
θ = 1/√(5) = √(5)/5 rad
Direction Angles
Given a = <a1, a2, a3>
cos α =
cos β =
cos γ =
Example
Find the direction angles of the vector a:
Given a = 3i – j + 4k
|a| = √(9+1+16) = √(26)
cos α = 3/√(26)  α = cos-1(3/√(26)) = 0.94 rad
cos β = -1/√(26)  β = cos-1(-1/√(26)) = 1.8 rad
cos γ = 4/√(26)  γ = cos-1(4/√(26)) = 0.67 rad
Projections
Scalar projection of b onto a:
compa b =
Vector projection of b onto a:
proja b = (
) =
Example
Find the scalar and vector projection of b onto a:
Given a = <-1, -3> and b = <3, -3>
a · b = (-1)(3) + (-3)(-3) = -3 + 9 = 6
|a| = √(1+9) = √(10)
|b| = √(9+9) = √(18) = 3√(2)
compa b = a · b/|a| = 6/√(10)
proja b = a · b/|a|2*a = 6/10*a = <-3/5, -9/5>
Application Example 1
Problem: A cart is pulled a distance of 50m along a horizontal
path by a constant force of 25 N. The handle of the cart is
pulled at an angle of 60° above the horizontal. Find the work
done by the force.
Solution:
F and d are force and displacement vectors
W = F · d = |F||d|cosθ = (25)(50)cos(60°)
W = 625 J
Application Example 2
Problem: Given a constant vector field F = 7i + 3j – k find the
work done from point P(5,3,-4) to the point Q(1,4,-7)
Solution:
d = <1 – 5, 4 – 3, -7 – (-4)> = <-4, 1, -3>
W = F · d = <7,3,-1> · <-4,1,-3>
= (7)(-4) + (3)(1) + (-1)(-3) = -28 + 4 +3
W = -21 J
Vectors
Given a vector in any coordinate system, (rectangular,
cylindrical, or spherical) it is possible to obtain the
corresponding vector in either of the two other coordinate
systems
Given a vector A = Axax + Ayay + Azaz we can obtain A = Aρaρ +
AΦaΦ + Azaz and/or
A = Arar + AΦaΦ + Aθaθ
Rectangular Coordinate System
Define A = Axax + Ayay + Azaz as a rectangular vector where
each component is a function of x, y, and z
Cylindrical Coordinate System
Define A = Aρaρ + AΦaΦ + Azaz as a rectangular vector where
each component is a function of ρ, Φ, and z
Spherical Coordinate System
Define A = Arar + AΦaΦ + Aθaθ as a rectangular vector where
each component is a function of r, Φ, and θ
Rectangular to Cylindrical
Dot products of unit vectors in cylindrical and rectangular
coordinate systems
x = ρ cosΦ
y = ρ sinΦ
z=z
ax
ax
ax
ax
cos Φ
-sin Φ
0
ay
sin Φ
cos Φ
0
az
0
0
1
Rectangular to Spherical
Dot products of unit vectors in spherical and rectangular
coordinate systems
ar
Aθ
aΦ
ax
sin θ cos Φ
cos θ cos Φ
-sin Φ
ay
sin θ sin Φ
cos θ sin Φ
cos Φ
cos θ
-sin θ
0
x = r sinθ cosΦa
z
y = r sinθ sinΦ
z = r cosθ
Conversion
Given a rectangular vector A = Axax + Ayay + Azaz , we want to
find the vector in cylindrical coordinates A = Aρaρ + AΦaΦ + Azaz
To find any desired component of a vector, we take the dot
product of the vector and a unit vector in the desired
direction.
Aρ = A · aρ and A Φ = A · aΦ
Example 1
Express the vector F = 4ax -2ay +8az in cylindrical coordinates:
Fρ = F · aρ = 4(ax · aρ) – 2(ay · aρ) + 8(az · aρ)
= 4(cosΦ) – 2(sinΦ) + 8(0)
= 4cosΦ – 2sinΦ
FΦ = F · aΦ = 4(ax · aΦ) – 2(ay · aΦ) + 8(az · aΦ)
= 4(-sinΦ) – 2(cosΦ) + 8(0)
= -4sinΦ – 2cosΦ
Fz = 8
F = (4cosΦ – 2sinΦ)aρ + (-4sinΦ – 2cosΦ)aΦ + 8az
Example 1 (cont.)
Evaluate F given ρ = 2.5, Φ = 0.7, z = 1.5:
F = (4cosΦ – 2sinΦ)aρ + (-4sinΦ – 2cosΦ)aΦ + 8az
Fρ = 4cosΦ – 2sinΦ = 4cos(0.7) – 2sin(0.7) = 1.77
FΦ = -4sinΦ – 2cosΦ = -4sin(0.7) – 2cos(0.7) = -4.1
Fz = 8
F = 1.77aρ – 4.1aΦ + 8az
Example 2
Given a vector field E = xax + yay + zaz, convert to cylindrical and
spherical coordinates:
Eρ = E · aρ = x(ax · aρ) + y(ay · aρ) + z(az · aρ)
= ρcosΦ(cosΦ) – ρsinΦ(sinΦ) + z(0)
= ρcos2Φ + ρsin2Φ = ρ(cos2Φ + sin2Φ) = ρ
EΦ = E · aΦ = x(ax · aΦ) + y(ay · aΦ) + z(az · aΦ)
= ρcosΦ(-sinΦ) + ρsinΦ(cosΦ) + z(0)
= -ρcosΦsinΦ + ρcosΦsinΦ = 0
Ez = z
E = ρaρ + zaz (cylindrical)
Example 2 (cont.)
Express E in spherical coordinates:
Er = E · ar = x(ax · ar) + y(ay · ar) + z(az · ar)
= x(sinθcosΦ) + y(sinθsinΦ) + z(cosθ)
= rsin2θcos2Φ + rsin2θsin2Φ + rcos2θ
= rsin2θ(cos2Φ + sin2Φ) + rcos2θ = rsin2θ + rcos2θ
= r(sin2θ + cos2θ) = r
Eθ = E · aθ = x(ax · aθ) + y(ay · aθ) + z(az · aθ)
= x(cosθcosΦ) + y(cosθsinΦ) + z(-sinθ)
= rsinθcosθcos2Φ + rcosθsinθsin2Φ – rsinθcosθ
= rsinθcosθ(cos2Φ + sin2Φ) – rsinθcosθ
= rsinθcosθ – rsinθcosθ = 0
Example 2 (cont.)
EΦ = E · aΦ = x(ax · aΦ) + y(ay · aΦ) + z(az · aΦ)
= x(-sinΦ)+ y(cosΦ)
= -rsinθsinΦcosΦ + rsinθsinΦcosΦ = 0
E = r ar (spherical)
References
• Calculus – Stewart 6th Edition, Section 13.3 “The Dot Product”